Question

Use the "mixed partials" check to see if the following differential equation is exact. If it is exact find a function ????(x,y) whose differential, ????????(x,y) gives the differential equation. That is, level curves ????(x,y)=???? are solutions to the differential equation: ????y????x=−3x4+2y−2x−4y2 First rewrite as ????(x,y)????x+????(x,y)????y=0 where ????(x,y)= , and ????(x,y)= . If the equation is not exact, enter not exact, otherwise enter in ????(x,y) as the solution of the differential equation here =????.

Answer 1

Answer:

The differential equation is exact.

So the solution of the differential equation is $F(x,y)=\frac{4}{5}x^5 +3yx +\frac{2y^5}{5}$

Step-by-step explanation:

So the question is to check whether or the given differential equation

    $\frac{dy}{dx} =\frac{-4x^{4}-3y}{3x+2y^{4}}$

is exact

Step One : Put the differential equation in standard form

To check whether it is exact or not we are going to put it in form by cross-multiplication

  $(3x + 2y^4)dy = (-4x^4 -3y)dx$

Since all the terms in the dx side are negative we are going to move them to the other side to get everything on the same side so we have

$(4x^4 + 3y)dx+ (3x+2y^4)dy = 0$

and this is in the standard form which is

   $Mdx +Ndy =0$

Step Two: Take the partial differentiation of the standard form and compare it there are equal

       $\frac{\delta M}{\delta y} = 3$

        $\frac{\delta N}{\delta x} = 3$

Since their partial differential is equal then the differential equation is exact

Step Three : Solve the differential equation

Now that we know it is exact how do we solve it well we are going to take

of $Mdx$ which is

       $\int ({4x^4 +3y}) \ dx = \frac{4x^5}{5}+3yx + C(y) -------(1)$  

The reason for the  C(y) is because the equation is been integrated with respect to x so any function in y will have derivative 0 with respect to x so we are going to be left with some arbitrary function in y

Now to figure this arbitrary function C(y) we are going to take the derivative of the function with respect to y and then compare it to what it is suppose to be which is N

   $\frac{\delta (\frac{4}{5}x^5 +3yx +C(y) )}{\delta y} = 3x +C'(y) = N$

and $N = 3x + 2y^4$

So we have

=>    $3x +C '(y) = 3x + 2y^4$

=>    $C'(y) = 2y^4$

Taking integral of both sides

$g(y) = \frac{2y^5}{5} +C$

put it back into equation one we have

   $F(x,y)=\frac{4}{5}x^5 +3yx +\frac{2y^5}{5}$